Question

Find the sum of the first 25 terms of an A.P. whose nth term is given by tₙ = 2 – 3n.


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Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt Find\: the\: sum\: of\: the\: first \:25 \:terms\: of \:an \:A.P.\\\tt whose \:n\:th\: terminal\: is \:given \:by \:t_n= 2 – 3n.$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

• $\bf n\:th \:term\: of\: an\: A.P=2-3n$

$\sf \bf {\boxed {\mathbb {TO\:FIND}}}$

• $\bf Sum\: of\: the\: first \:25 \:terms \:of \:this\: A.P$

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

${\pink {\underline {\pmb {\bf {Sum\: of\: the\: first \:25 \:terms \:of \:this\: A.P}}}}}$

${\leadsto{\orange{\sf {When \:n=1}}}}$

$\bf \implies t_1=2-3\big(1\big)$

$\bf \implies t_1=2-3$

$\implies {\blue {\boxed {\boxed {\purple {\sf {t_1=-1}}}}}}$

${\leadsto{\orange{\sf {When \:n=25}}}}$

$\bf \implies t_1=2-3\big(25\big)$

$\bf \implies t_1=2-75$

$\implies {\blue {\boxed {\boxed {\purple {\sf {t_1=-73}}}}}}$

${\blue {\boxed {\boxed {\boxed {\green {\pmb {S_n=\dfrac{n}{2}\big[t+t_n\big]}}}}}}}$

• $\sf S_n=sum\: of\: the \:n\: terms$
• $\sf t=1\:st\:term\:of\: the \:A.P$
• $\sf t_n=n\:th\:term\:of\: the \:A.P$

${\underbrace {\overbrace {\orange {\pmb {Substitute \:the \:values}}}}}$

$\bf \implies S_{25}=\dfrac{25}{2}\Big[-1+\big(-73\big)\Big]$

$\bf \implies S_{25}=\dfrac{25}{2}\Big[-1-73\Big]$

$\bf \implies S_{25}=\dfrac{25}{2}\times -74$

$\bf \implies S_{25}=\dfrac{25}{\cancel{2}}\times \cancel{-74}$

$\bf \implies S_{25}=25\times - 37$

$\implies {\blue {\boxed {\boxed {\purple {\mathfrak {S_{25}=-925}}}}}}$

${\underbrace {\red {\underline {\red {\overline {\red {\pmb {\sf {{\therefore} The\:sum\: of\: the\: first \:25 \:terms \:of \:this\: A.P\:is\:-925}}}}}}}}}$

___________________________________

$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

${\pink {\bf {Formulas}}}$

$\sf a_n=a+(n-1)d$

$\sf S_n=\dfrac{n}{2}[t+t_n]$

$\sf S_n=\dfrac{n}{2}\big[2t+(n-1)d\big]$

Answer 2

Answer:

• -925 is the sum of first 25 terms of the A.P whose nth term is given by tₙ = 2-3n.

Step-by-step explanation:

Given:

• nth term of the A.P tₙ = 2-3n.

To find:

• Sum of the first 25 terms.

Solution:

Let's find the first (initial terms of the A.P)

n = 1

• 2 - 3(1) = -1 (a1)

n = 2

• 2 - 3(2) = -4 (a2)

n = 3

• 2 - 3(3) = -7 (a3)

Clearly, common difference:

• a2 - a1
• -4 - (-1)
• -3

Sum of n terms of an A.P is given by:

⇒ Sₙ = n/2[2(a) + (n-1)d]

• S₂₅ = 25/2[2(-1) + (25-1)-3]
• S₂₅ = 25/2[-2 + 24(-3)]
• S₂₅ = 25/2[-2 - 72]
• S₂₅ = 25/2(-74)
• S₂₅ = -925

∴ The sum of first 25 terms of the A.P whose nth term is given by tₙ = 2-3n is -925.